A) 25 m, 10 m
B) 13 m, 27 m
C) 22 m, 13 m
D) 17 m, 20 m
Correct Answer: C
Solution :
It is the case of Booster angle where reflected and refracted ray both become polarized. So, \[{{\mu }_{1}}=\tan {{i}_{B}}\] \[\frac{4}{3}=\tan {{i}_{B}}\] From triangle ABC, \[\tan (90-{{i}_{B}})=\frac{BC}{AB}\] \[\cot \,{{i}_{B}}=\frac{10}{x}\] \[\frac{3}{4}=\frac{10}{x}\] \[x=\frac{40}{3}=13\,m\] From triangle AEF, \[\tan (90-{{i}_{B}})=\frac{y}{50-x}\] \[\cot {{i}_{B}}=\frac{y}{50-13}\] \[\frac{3}{4}=\frac{y}{37}\] \[y=\frac{3\times 37}{4}\] \[y=27\,m\]You need to login to perform this action.
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